Hey I was wondering how the power rating of chokes works. For instance a tube amp Choke rated for say 600 volts and 200ma's. Is that really based on a wattage rating, so at a lower voltage it would handle more current, or is the current rating fixed.
I want to build a pi filter for a Class AB solid state amplifier and the current demands are greater than 200ma, obviously. However the voltage is 63 volts, much lower, so I wondered if some of these standerd off the shelf units owuld work for my needs.
If not, I have been asking some places for quotes on custom made ones designed for my needs. Namely 100 volts, 5H, and 10 amps. Does that make sense, or am I specing them wrong?
I want to build a pi filter for a Class AB solid state amplifier and the current demands are greater than 200ma, obviously. However the voltage is 63 volts, much lower, so I wondered if some of these standerd off the shelf units owuld work for my needs.
If not, I have been asking some places for quotes on custom made ones designed for my needs. Namely 100 volts, 5H, and 10 amps. Does that make sense, or am I specing them wrong?
For typical tube amps, chokes have high inductance ratings because of lower currents. If you really want a choke for a S.S. amp that needs, say 4 amps, you can have a much lower rated choke. Chances are there are not very many 5H 4 amp chokes readily available because the windings will be thicker due to increased current requirement and the core size will be huge.
Rule of thumb is for choke input power supplies: the lower current that you need, the higher inductance you will need...and vice versa. This is because to allow the same ripple voltage buffer (for lack of better words), the L and di/dt will change (go up, and go down, or oppositely) in the standard inductor equation V(t) = L*[di(t)/dt], where V(t) is voltage across the inductor and i(t) is current through the inductor.
You can also consider the inductor as an impedance in the frequency domain. With tube amps, a typical load of tubes might present a high voltage supply with say, 1kohm. Conversely, a S.S. amp might present it's supply with a typical load of 20 ohms. An inductor represents an impedance of j*w*L. Since w is constant at 2*pi*60, L determines impedance. Say that we want a filter impedance of %50 of the load value (500ohms for the tube amp and 10 ohms for the SS amp...remeber this is only AC impedance, power is only dissipated int the winding resistance (plus a little core loss)). This would mean an inductance of 1.3 Henries for the tube amp but only 27mHenries for the S.S. amp.
In other words, look for a much smaller inductor and they will likely have higher current ratings. Don't worry too much about the voltage ratings unless you are designing for a tube amp. Operating an inductor past its current rating will saturate the core and cause inductance to drop. It also risks overheating the windings and fire (if way past the current rating).
hope this helps,
Rule of thumb is for choke input power supplies: the lower current that you need, the higher inductance you will need...and vice versa. This is because to allow the same ripple voltage buffer (for lack of better words), the L and di/dt will change (go up, and go down, or oppositely) in the standard inductor equation V(t) = L*[di(t)/dt], where V(t) is voltage across the inductor and i(t) is current through the inductor.
You can also consider the inductor as an impedance in the frequency domain. With tube amps, a typical load of tubes might present a high voltage supply with say, 1kohm. Conversely, a S.S. amp might present it's supply with a typical load of 20 ohms. An inductor represents an impedance of j*w*L. Since w is constant at 2*pi*60, L determines impedance. Say that we want a filter impedance of %50 of the load value (500ohms for the tube amp and 10 ohms for the SS amp...remeber this is only AC impedance, power is only dissipated int the winding resistance (plus a little core loss)). This would mean an inductance of 1.3 Henries for the tube amp but only 27mHenries for the S.S. amp.
In other words, look for a much smaller inductor and they will likely have higher current ratings. Don't worry too much about the voltage ratings unless you are designing for a tube amp. Operating an inductor past its current rating will saturate the core and cause inductance to drop. It also risks overheating the windings and fire (if way past the current rating).
hope this helps,
Thanks that was very helpful. Now that I know this I believe I could get away with an off the shelf part from Hammond, or possibly a much cheaper custom made one.
Now, is there any issue with the current rating? In other words, should it be rated for the entire possible draw of the amp, or lower?
Now, is there any issue with the current rating? In other words, should it be rated for the entire possible draw of the amp, or lower?
The current rating is going to depend on the wire gauge used in the choke, since a tube amp uses higher voltage at lower current they can use relatively thin wire, whereas for a high current SS amp you'd need a choke with much heavier wire.
The DC power dissipated in the choke will be [I squared * R] where R is the DC resistance of the choke and I is the DC current, so if you had a 50 ohm choke at 200mA it would only be 2 watts but at 2A it would be 200W!
The DC power dissipated in the choke will be [I squared * R] where R is the DC resistance of the choke and I is the DC current, so if you had a 50 ohm choke at 200mA it would only be 2 watts but at 2A it would be 200W!
well I understood the first part already, what I wondering was how to choose the current value. I was reading that in an LC where there is no C after the bridge, the Inductor must be able to handle the full current swings, but with a CLC that isn't the case, however this was discussing it around tube amps. I've seen all this discussed for solid state amps as well with regard to Class A amps such as Pass designs, but again he is using inductors rated far below the total draw of the amp. I believe in one of his projects they are 20mh at 3amps. The amps power supply however can produce over 10 amps.
I don't know about a pi filter but for a simple choke input filter
the formulia L (henries) =V/I (miliamps) x.88 so for a 20v
supply pulling 4A we need a inductor of at least 5mh. Now if
you use an iron core inductor you need a rting greater than
4A because somewhere right above the rated current the thing
will start to saturate. In this case an air core inductor might be
a good choice but you need to watch the wattage disipated
in the choke.
the formulia L (henries) =V/I (miliamps) x.88 so for a 20v
supply pulling 4A we need a inductor of at least 5mh. Now if
you use an iron core inductor you need a rting greater than
4A because somewhere right above the rated current the thing
will start to saturate. In this case an air core inductor might be
a good choice but you need to watch the wattage disipated
in the choke.
If I want to figure out what I need for a supply that is 45v and 10 amps, I came up with one of at least 4mh and I was planning on using one that was rated for 10-25mh depending on what I can get.
Triad Magnetics makes some affordable ones, like 50 bucks each, so I could get two of them and be in good shape. Hmm would using two smaller ones in a CLCLC filter be better than one bigger one in a CLC filter?
I looked at an LC filter and it didn't seem to be as effective at reducing ripple as a CLC filter, but I was reading that they regulate voltage better? I would have thought that would go hand in hand, but I dont know, I'm just getting to understand this stuff.
Triad Magnetics makes some affordable ones, like 50 bucks each, so I could get two of them and be in good shape. Hmm would using two smaller ones in a CLCLC filter be better than one bigger one in a CLC filter?
I looked at an LC filter and it didn't seem to be as effective at reducing ripple as a CLC filter, but I was reading that they regulate voltage better? I would have thought that would go hand in hand, but I dont know, I'm just getting to understand this stuff.
Hi,
the output from a choke input filter is about Vdc=0.9Vac.
The output for a capacitor input filter is about Vdc=1.4Vac.
The inductor for a choke input filter should be designed to not overheat when on high demand and provide sufficient inductance to ensure correct regulation to achieve that 0.9Vac value.
There is a problem.
If the current demand falls below the lower limit for choke regulation, the voltage on the output starts to rise and will approach the 1.4Vac when the current demand is very low.
ClassAB amplifiers with a high ratio between quiescent current and Ipeak may fall into that category of PSU supply voltage swinging between 1.4 and 0.9 times Vac.
Choosing a high output bias design may avoid this problem.
Understand the design implications and take the necessary steps to ensure your supply voltage does not jump by that 50% unexpectedly.
BTW,
the reason you rarely find choke input filters in solid state amplifiers is that the choke will probably be as big as and as expensive as the main transformer and could be even bigger/more costly.
the output from a choke input filter is about Vdc=0.9Vac.
The output for a capacitor input filter is about Vdc=1.4Vac.
The inductor for a choke input filter should be designed to not overheat when on high demand and provide sufficient inductance to ensure correct regulation to achieve that 0.9Vac value.
There is a problem.
If the current demand falls below the lower limit for choke regulation, the voltage on the output starts to rise and will approach the 1.4Vac when the current demand is very low.
ClassAB amplifiers with a high ratio between quiescent current and Ipeak may fall into that category of PSU supply voltage swinging between 1.4 and 0.9 times Vac.
Choosing a high output bias design may avoid this problem.
Understand the design implications and take the necessary steps to ensure your supply voltage does not jump by that 50% unexpectedly.
BTW,
the reason you rarely find choke input filters in solid state amplifiers is that the choke will probably be as big as and as expensive as the main transformer and could be even bigger/more costly.
Could you further explain what you mean Andrew, what your saying would explain a problem I was having with my models. I'm basicly not sure how to "do the math" to ensure that doesn't happen.
When I modeled my power supplies on PSUD2 sometimes it would show me output that was more like 150 volts, other times it gave me the expected 60 volts. I thought it might be a glitch in the program, because if I shut it down and rerun it would give me what I expected.
I know the Inductors are typically large and pricey, but based on other information I heard, It seemed I could get away with a 5-10mh inductor that handles around 10 amps. Triad magnetics has some which are, I believe, designed for my purpose, and will cost around 55 dollars each or so. How do I figure out the minimum current that must be drawn in order for the inductors to maintain regulation.
Also, did I gather correctly that an LC will be better than a CLC? Thanks.
When I modeled my power supplies on PSUD2 sometimes it would show me output that was more like 150 volts, other times it gave me the expected 60 volts. I thought it might be a glitch in the program, because if I shut it down and rerun it would give me what I expected.
I know the Inductors are typically large and pricey, but based on other information I heard, It seemed I could get away with a 5-10mh inductor that handles around 10 amps. Triad magnetics has some which are, I believe, designed for my purpose, and will cost around 55 dollars each or so. How do I figure out the minimum current that must be drawn in order for the inductors to maintain regulation.
Also, did I gather correctly that an LC will be better than a CLC? Thanks.
Hi,
it took me a bit of searching to find a link that included the formulae.
hope this helps
http://www.linearmagnetics.com/Pages/LMC Engineering Information.html
para4.
I tried 50Vdc and 100mA min current and 5A continuous current and the formulae predicted 5.3H with 2r0 total inductor and secondary resistance. If secondary resistance = 0r5 then inductor resistance will need to be 1r5. This then needs a transformer with Vac=67.1Vac.
Try some numbers in psud2 for various Rload values and examine Vdc and output ripple.
Note, that simply changing Rmax from 5k (100mA) to 10k (50mA) changes the required inductor to 10.6H There's that problem, low bias current and risk of overvoltage on the amp supply rails. Psud should be able to predict the rise in voltage with falling bias current and you can decide on minimum inductance (budget) to meet the amplifier design needs.
I do wish folk would at least try to do their own research (my rant)
it took me a bit of searching to find a link that included the formulae.
hope this helps
http://www.linearmagnetics.com/Pages/LMC Engineering Information.html
para4.
I tried 50Vdc and 100mA min current and 5A continuous current and the formulae predicted 5.3H with 2r0 total inductor and secondary resistance. If secondary resistance = 0r5 then inductor resistance will need to be 1r5. This then needs a transformer with Vac=67.1Vac.
Try some numbers in psud2 for various Rload values and examine Vdc and output ripple.
Note, that simply changing Rmax from 5k (100mA) to 10k (50mA) changes the required inductor to 10.6H There's that problem, low bias current and risk of overvoltage on the amp supply rails. Psud should be able to predict the rise in voltage with falling bias current and you can decide on minimum inductance (budget) to meet the amplifier design needs.
I do wish folk would at least try to do their own research (my rant)
I tried to do my own research, but being unclear on the concept you were talking about makes it hard to know what to search for. Thank you though, its much appreciated.
You are coming up with numbers that once again are making a pi filter look unreasonable. Did you mean to say 5-10mh and not henries when you quoted numbers? I mean, to give you an idea, a 5mh inductor that handles 10 amps at 100 volts is around 15" by 15" by 15" or so. That is based on quotes from various sources, and would be priced at around 2500-3500 dollars each. In looking further into what sorts of inductors were used in solid state amps running lower voltages, the information I was getting was suggesting mh inductors not h inductors, which made is plausible. If you look at the Cello amps, Musical Fidelity, etc they all use inductors similar in size to the transformer, which implies they are no more than 20mh or so, 100 maybe for the cello's.
You are coming up with numbers that once again are making a pi filter look unreasonable. Did you mean to say 5-10mh and not henries when you quoted numbers? I mean, to give you an idea, a 5mh inductor that handles 10 amps at 100 volts is around 15" by 15" by 15" or so. That is based on quotes from various sources, and would be priced at around 2500-3500 dollars each. In looking further into what sorts of inductors were used in solid state amps running lower voltages, the information I was getting was suggesting mh inductors not h inductors, which made is plausible. If you look at the Cello amps, Musical Fidelity, etc they all use inductors similar in size to the transformer, which implies they are no more than 20mh or so, 100 maybe for the cello's.
After freaking out about this issue I realized that we are only talking a small bit of voltage change here that is all within the working range of the amplifier. I imagine its not the ideal from an output standpoint as you would want the higher working voltage as demand increases, but I dont see it as being damaging. Am I right about that? I mean, the amp can handle up to 75 volts, it will put out around 62 with low current draw and 57 with the higher current draw. Even if I increase the voltage up I see it not being a big deal. I might have the transformer resistance wrong though to get the effect you are talking about, I'm not sure what the resistance would be.
Hi,
I cannot check my arithmetic at the moment, but could I have lost 10^3 in my calculation?
I saw 5.3H as the answer for a 5k minimum load. I used 5k as the Rmax. This is equivalent to a 10mA load on 50Vdc PSU. Oops there's my first mistake. 500r should be Rmax for 100mA quiescent current using a 50Vdc supply and that brings the inductor down to 530mH .
Have you tried your own numbers yet?
If the output of the choke input PSU is swinging around 57 to 62V that completely defeats the intended advantage this LC is supposed to have. You are budgetting to improve regulation not buy in worse performance.
As far as I can see transformer resistance is not critical. MUCH more important is the ratio of minimum output current to maximum output current. Very short term peaks can be sourced from the last C in the PSU, then the energy in the inductor recharges the C ready for the next peak. In the meantime the LC meets the more normal current draw.
I do not know what quiescent current the MF choke filtered amps use, but having seen how big the sinks are in the few models that choose this style of PSU, I guess they have adopted a very high bias design and that would complement a smaller inductor.
I cannot check my arithmetic at the moment, but could I have lost 10^3 in my calculation?
I saw 5.3H as the answer for a 5k minimum load. I used 5k as the Rmax. This is equivalent to a 10mA load on 50Vdc PSU. Oops there's my first mistake. 500r should be Rmax for 100mA quiescent current using a 50Vdc supply and that brings the inductor down to 530mH .
Have you tried your own numbers yet?
If the output of the choke input PSU is swinging around 57 to 62V that completely defeats the intended advantage this LC is supposed to have. You are budgetting to improve regulation not buy in worse performance.
As far as I can see transformer resistance is not critical. MUCH more important is the ratio of minimum output current to maximum output current. Very short term peaks can be sourced from the last C in the PSU, then the energy in the inductor recharges the C ready for the next peak. In the meantime the LC meets the more normal current draw.
I do not know what quiescent current the MF choke filtered amps use, but having seen how big the sinks are in the few models that choose this style of PSU, I guess they have adopted a very high bias design and that would complement a smaller inductor.
This is getting a bit 'out there.' Most even basic solid state amplifiers have excellent power supply rejection at 60/120Hz, unless they are class A. The only advantage i might see is using an LC filter is with the driver section. A lot of times they have a slightly separate supply anyway, usually higher to be able to drive MOSFETs to the PS rail with high Vgs voltages. The driver section might benefit from better DC power, depending on topology.
Think about it...how often have you heard hum problems from a solid state amp?? If there is...it is usually getting in the signal path from outside sources or a ground loop problem.
So I guess the question would be...do you have a schematic of the amp designed for? Can you isolate the driver supply (usually you can) to use your proposed LC filter. Using choke input may drop the voltage too much to be able to swing the output transistors far enough though...
Think about it...how often have you heard hum problems from a solid state amp?? If there is...it is usually getting in the signal path from outside sources or a ground loop problem.
So I guess the question would be...do you have a schematic of the amp designed for? Can you isolate the driver supply (usually you can) to use your proposed LC filter. Using choke input may drop the voltage too much to be able to swing the output transistors far enough though...
Thank's for the pics, I still feel that the chokes even in that amp aren't that large. We already caught one math problem, the one causing a bit of it, so at least I know it should be under 1H to fit this issue. I think I will actually have more like 500ma quiscent current total draw, since there will be three amplifiers on one common power supply. That will help further bring down the need.
To see if I'm clear, would going with a CLC instead of LC completely eliminate the need for the larger inductor because of the swinging current draw?
To see if I'm clear, would going with a CLC instead of LC completely eliminate the need for the larger inductor because of the swinging current draw?
yeah but now that you have told me the current saturation, that reinforces even more to me that the amp isn't using a very large inductor value, which is what I meant when I said large. Again, when I asked for a quote on a 1H inductor at 15 amps it was over twice as large as that one appears, being 18 inches in diameter and 12 inches high. An Air gapped inductor rated for 5H and 10 amps was 15" by 15" by 15", and the cost on both was over 3 grand. I'm currently working with Toroid corporation and they are telling me that the 20mh rated to saturate at 25amps is going to around 8" in diameter by 6" high, so for the sake of arguement, I bet the Bravo uses something more like that. Yes thats very large in physical size, yes that is handeling some huge current, but the MH rating is still quite small compared with what has been discussed, and what you see typically in higher voltage designs.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.